Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/646.html

646. Maximum Length of Pair Chain (Medium)

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion.

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:

Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4]

Note:

1. The number of given pairs will be in the range [1, 1000].

Related Topics:
Dynamic Programming

Similar Questions:

• Longest Increasing Subsequence (Medium)
• Increasing Subsequences (Medium)

Solution 1. DP

First sort the array in ascending order.

Let dp[i] be the length of maximum chain formed using a subsequence of A[0..i] where A[i] must be used.

dp[i] = max(1, max(1 + dp[j] | pair j can go after pair i ))

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int findLongestChain(vector<vector<int>>& A) {
        sort(begin(A), end(A));
        int N = A.size();
        vector<int> dp(N, 1);
        for (int i = 1; i < N; ++i) {
            for (int j = 0; j < i; ++j) {
                if (A[j][1] >= A[i][0]) continue;
                dp[i] = max(dp[i], 1 + dp[j]);
            }
        }
        return *max_element(begin(dp), end(dp));
    }
};

Solution 3. Interval Scheduling Maximization (ISM)

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int findLongestChain(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
        int e = INT_MIN, ans = 0;
        for (auto &v : A) {
            if (e >= v[0]) continue;
            e = v[1];
            ++ans;
        }
        return ans;
    }
};

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int findLongestChain(int[][] pairs) {
          Arrays.sort(pairs, new Comparator<int[]>() {
              public int compare(int[] pair1, int[] pair2) {
                  if (pair1[0] != pair2[0])
                      return pair1[0] - pair2[0];
                  else
                      return pair1[1] - pair2[1];
              }
          });
          int length = pairs.length;
          int[] dp = new int[length];
          Arrays.fill(dp, 1);
          for (int i = 1; i < length; i++) {
              int[] curPair = pairs[i];
              for (int j = 0; j < i; j++) {
                  int[] prevPair = pairs[j];
                  if (prevPair[1] < curPair[0])
                      dp[i] = Math.max(dp[i], dp[j] + 1);
              }
          }
          int max = 1;
          for (int num : dp)
              max = Math.max(max, num);
          return max;
      }
  }
  
  ############
  
  class Solution {
      public int findLongestChain(int[][] pairs) {
          Arrays.sort(pairs, Comparator.comparingInt(a -> a[1]));
          int ans = 0;
          int cur = Integer.MIN_VALUE;
          for (int[] p : pairs) {
              if (cur < p[0]) {
                  cur = p[1];
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
  // Time: O(N^2)
  // Space: O(N)
  class Solution {
  public:
      int findLongestChain(vector<vector<int>>& A) {
          sort(begin(A), end(A));
          int N = A.size();
          vector<int> dp(N, 1);
          for (int i = 1; i < N; ++i) {
              for (int j = 0; j < i; ++j) {
                  if (A[j][1] >= A[i][0]) continue;
                  dp[i] = max(dp[i], 1 + dp[j]);
              }
          }
          return *max_element(begin(dp), end(dp));
      }
  };
  

• class Solution:
      def findLongestChain(self, pairs: List[List[int]]) -> int:
          ans, cur = 0, -inf
          for a, b in sorted(pairs, key=lambda x: x[1]):
              if cur < a:
                  cur = b
                  ans += 1
          return ans
  
  ############
  
  class Solution(object):
    def findLongestChain(self, pairs):
      """
      :type pairs: List[List[int]]
      :rtype: int
      """
      tails = []
      for start, end in sorted(pairs):
        idx = bisect.bisect_left(tails, start)
        if idx == len(tails):
          tails.append(end)
        else:
          tails[idx] = min(tails[idx], end)
      return len(tails)
  
  

• func findLongestChain(pairs [][]int) int {
  	sort.Slice(pairs, func(i, j int) bool {
  		return pairs[i][1] < pairs[j][1]
  	})
  	ans, cur := 0, math.MinInt32
  	for _, p := range pairs {
  		if cur < p[0] {
  			cur = p[1]
  			ans++
  		}
  	}
  	return ans
  }
  

• function findLongestChain(pairs: number[][]): number {
      pairs.sort((a, b) => a[1] - b[1]);
      let res = 0;
      let pre = -Infinity;
      for (const [a, b] of pairs) {
          if (pre < a) {
              pre = b;
              res++;
          }
      }
      return res;
  }
  
  

• impl Solution {
      pub fn find_longest_chain(mut pairs: Vec<Vec<i32>>) -> i32 {
          pairs.sort_by(|a, b| a[1].cmp(&b[1]));
          let mut res = 0;
          let mut pre = i32::MIN;
          for pair in pairs.iter() {
              let a = pair[0];
              let b = pair[1];
              if pre < a {
                  pre = b;
                  res += 1;
              }
          }
          res
      }
  }
  
  

All Problems

All Solutions